Problem: Factor the following expression: $x^2 + 10x + 21$
Solution: When we factor a polynomial, we are basically reversing this process of multiplying linear expressions together: $ \begin{eqnarray} (x + a)(x + b) &=& xx &+& xb + ax &+& ab \\ \\ &=& x^2 &+& {(a + b)}x &+& {ab} \end{eqnarray} $ $ \begin{eqnarray} \hphantom{(x + a)(x + b) }&\hphantom{=}&\hphantom{ xx }&\hphantom{+}&\hphantom{ (a + b)x }&\hphantom{+}& \\ &=& x^2 & +& {10}x& +& {21} \end{eqnarray} $ The coefficient on the $x$ term is $10$ and the constant term is $21$ , so to reverse the steps above, we need to find two numbers that add up to $10$ and multiply to $21$ You can try out different factors of $21$ to see if you can find two that satisfy both conditions. If you're stuck and can't think of any, you can also rewrite the conditions as a system of equations and try solving for $a$ and $b$ $ {a} + {b} = {10}$ $ {a} \times {b} = {21}$ The two numbers $3$ and $7$ satisfy both conditions: $ {3} + {7} = {10} $ $ {3} \times {7} = {21} $ So we can factor the expression as: $(x + {3})(x + {7})$